All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > General
Four Way Smarties (Posted on 2004-11-05) Difficulty: 3 of 5
You and four other people (who coincendentally are all smarties) are in late testing room where you will take your test where there is a 6 by 6 grid of equally spaced desks with chairs in the same relative spot.

You go into the room after all four smarties have chosen their location. You have a test taking policy where you always want to sit at the midpoint between two smarties. The smarties in the room with you feel the exact opposite way, so their arrangement is always such that no smartie is at the midpoint of two other smarties

However, depending on where the smarties are sitting, you may not be able to sit at the midpoint since in all cases it would always be where there is no chair and desk. (There is a strict no moving desks or chairs rule too.) How many ways could the current 4 smarties sit such that you couldn't sit at the midpoint of two smarties if reflections and rotations count as well?

How many ways could you not find where you want to sit if there were 5 smarties other than you and reflections and rotations count as well?

See The Solution Submitted by Gamer    
Rating: 4.2000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Part B (Full Solutions--with six smarties (including myself) | Comment 9 of 22 |
(In reply to re: Part B (Correction and Full Solutions--Possibly) by Osi)

4 smarties to sit in thegrid

This is quite tricky because you can position the smarties anywhere and yield a possible number of ways quite different from others. here are are two ways of positioning.

Way 1: All four smarties could position themselves in the middlemost grids--2 by 2 adjoining grids in the middle. For every sit one smartie takes, 9 are eliminated for the upcoming smartie--that is the smartie's own sit and 8 others that prevent possible middle positions. Again, there are 4 orders of sitting. So the number of ways this could happen is 4*36*27*18*9 = 629,856.

Way 2: Now suppose you wanted a larger number of ways, all smarties would sit at the corners meaning 4 seats would eliminated for other subsequent smarties. Therefore, number sittings possible, with the 4 possible orders taken into consideration, we have 4*36*32*28*24 = 3096576 ways. This is larger than the the solution above. It makes sense to pick the larger number, therefore, I conclude that the maximum number of ways 4 smarties could sit, subject to the constraints of the question posed--no middle smarties--is 3,096,576 ways. 

Question Solutions:

1). In both solutions above, 20 sits are possible for the fifth smartie so that there are still no middle smarties. In method one above, the 20 sits are all around the periphery of the of 6 by 6 grid. In the second method yielding more number of ways, the 20 sits run around the 2 by 2 middle grids. Multiplying the two results above by 20, we reach  12,597,120 ways and 61,931,520 ways respectively. Again we pick the maximum number and 5 smarties can sit in 61,931,520 ways

2) For six smarties, that is 5 smarties including myself, you have to multiply the number above by 19. If the 5th smartie takes a seat as described in (1) above, I can have 19 seats such that the constraint in the question posed is satisfied. It would seem as it is 16 ways I can seat, however seating near the 5th smartie doesn't violate the no-middle-man constraint. Therefore, I have 19 ways of taking my seats. Hence answer is 61,931,520*19 = 1,176,698,880.


  Posted by Osi on 2004-11-05 18:19:41
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (5)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (7)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information